Question

In Problems 53-56, use the properties of limits and the facts that

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\sin x}{x}=1 \\ \underset{x\to 0}{\lim }\frac{\cos x-1}{x}=0 \\ \underset{x\to 0}{\lim }\sin x=0 \\ \underset{x\to 0}{\lim }\cos x=1 \end{gathered}$$

$\underset{x\to 0}{\lim }\frac{{\sin }^{2}x+\sin x(\cos x-1)}{x^2}$

Short answer

The limit is$1$

Step-by-step solution

Given information

Given function$\underset{x\to 0}{\lim }\frac{{\sin }^{2}x+\sin x(\cos x-1)}{x^2}$

Express in standard form and calculate

Calculating, we get

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{{\sin }^{2}x+\sin x(\cos x-1)}{x^2}=\frac{\sin x}{x}\left(\frac{\sin x}{x}\right)+\frac{\sin x}{x}\left(\frac{\cos x-1}{x}\right) \\ \underset{x\to 0}{\lim }\left(\frac{{\sin }^{2}x+\sin x(\cos x-1)}{x^2}\right)=\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\left(\frac{\sin x}{x}\right)+\frac{\sin x}{x}\left(\frac{\cos x-1}{x}\right)\right) \\ \underset{x\to 0}{\lim }\left(\frac{{\sin }^{2}x+\sin x(\cos x-1)}{x^2}\right)=\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)·\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)+\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)·\underset{x\to 0}{\lim }\left(\frac{\cos x-1}{x}\right) \\ \underset{x\to 0}{\lim }\left(\frac{{\sin }^{2}x+\sin x(\cos x-1)}{x^2}\right)=1\left(1\right)+1\left(0\right) \\ \underset{x\to 0}{\lim }\left(\frac{{\sin }^{2}x+\sin x(\cos x-1)}{x^2}\right)=1 \end{gathered}$$